Amines

The chapter on Amines in the 12th standard NCERT chemistry textbook covers the structure, classification, nomenclature, preparation, properties, and reactions of organic compounds containing nitrogen. Here’s a summary:

Key Concepts:

• Structure and Classification: Amines are derivatives of ammonia (NH3) where one or more hydrogen atoms are replaced by alkyl or aryl groups. They are classified as primary (RNH2), secondary (R2NH), or tertiary (R3N) depending on the number of alkyl/aryl groups attached to the nitrogen atom. Quaternary ammonium salts (R4N+ X-) have all four hydrogens replaced and carry a positive charge.
  
• Nomenclature: IUPAC names for amines involve identifying the longest carbon chain attached to the nitrogen and using “amine” as the suffix. Common names are often used, especially for simpler amines.
  
• Preparation: The chapter details several methods for preparing amines, including:
  
  • Reduction of nitro compounds, nitriles, and amides.
  • Ammonolysis of alkyl halides.
  • Gabriel phthalimide synthesis (for primary amines).
  • Hofmann degradation of amides.
• Physical Properties: Amines are typically gases or liquids at room temperature with a characteristic odor. Solubility decreases with increasing carbon chain length.
  
• Chemical Properties: Amines are basic due to the lone pair of electrons on the nitrogen atom. They react with acids to form salts. Key reactions include:
  
  • Reactions with acids (salt formation).
  • Reactions with alkyl halides (quaternary ammonium salt formation).
  • Reactions with aldehydes and ketones (Schiff base formation).
  • Reactions with acyl halides and anhydrides (amide formation).
  • Reaction with nitrous acid (different outcomes for primary, secondary, and tertiary amines).
  • Electrophilic aromatic substitution reactions (for aromatic amines).
• Basicity: The basicity of amines is influenced by factors like inductive effects, resonance, and steric hindrance. Generally, aliphatic amines are more basic than aromatic amines. The basicity order in aqueous solution is complex due to hydration effects.
  
• Distinction Tests: The chapter describes methods to distinguish between primary, secondary, and tertiary amines, such as the reaction with nitrous acid.
  
• Diazonium Salts: Aromatic primary amines react with nitrous acid to form diazonium salts, which are highly reactive intermediates.
  
• Reactions of Diazonium Salts: Diazonium salts undergo a variety of reactions, including:
  
  • Sandmeyer reaction.
  • Gattermann reaction.
  • Balz-Schiemann reaction.
  • Replacement by -H, -OH, -X, -CN.
  • Coupling reactions with phenols and aromatic amines.
• Importance of Amines: Amines are essential in various biological systems (e.g., neurotransmitters, hormones) and are used in the synthesis of dyes, drugs, and other organic compounds.

Exercise

1. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines. 

(i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2

(iv) (CH3)3 CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3

(vii)m-BrC6H4NH2

Ans : 

(i) (CH3)2CHNH2

• IUPAC Name: 2-Propanamine (or Isopropylamine)
• Classification: Primary amine

(ii) CH3(CH2)2NH2

• IUPAC Name: 1-Propanamine (or n-Propylamine)
• Classification: Primary amine

(iii) CH3NHCH(CH3)2

• IUPAC Name: N-Methyl-2-propanamine (or Methylisopropylamine)
• Classification: Secondary amine

(iv) (CH3)3CNH2

• IUPAC Name: 2-Methyl-2-propanamine (or tert-Butylamine)
• Classification: Primary amine

(v) C6H5NHCH3

• IUPAC Name: N-Methylaniline
• Classification: Secondary amine

(vi) (CH3CH2)2NCH3

• IUPAC Name: N-Ethyl-N-methyl ethanamine (or Ethylmethylamine)
• Classification: Tertiary amine

(vii) m-BrC6H4NH2

• IUPAC Name: 3-Bromoaniline (or m-Bromoaniline)
• Classification: Primary amine

2. Give one chemical test to distinguish between the following pairs of compounds:

(i)Methylamine and dimethylamine

(ii) Secondary and tertiary amines

(iii) Ethylamine and aniline

(iv) Aniline and benzylamine

(v) Aniline and N-Methylaniline.

Ans : 

3. Account for the following

(i) pKb of aniline is more than that of methylamine

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Ans : 

(i) 

• Basicity and pKb: A higher pKb value indicates a weaker base. Aniline is a much weaker base than methylamine.
• Resonance in Aniline: The lone pair of electrons on the nitrogen atom in aniline is delocalized into the benzene ring through resonance. This delocalization decreases the electron density on the nitrogen atom, making it less available to accept a proton. In other words, the lone pair is less available for donation to a proton.
• No Resonance in Methylamine: Methylamine, being an aliphatic amine, does not exhibit this resonance. The nitrogen’s lone pair is readily available for protonation.

(ii) 

• Hydrogen Bonding: Ethylamine, a small aliphatic amine, can form hydrogen bonds with water molecules due to the polar -NH2 group. This allows it to dissolve in water.
• Hydrophobic Character: Aniline, while also having a polar -NH2 group, has a large hydrophobic (water-repelling) phenyl group attached. The hydrophobic character of the phenyl ring outweighs the ability of the -NH2 group to form hydrogen bonds with water, making aniline much less soluble.

(iii) 

• Basic Nature of Amines: Methylamine, being a base, increases the pH of the aqueous solution.
• Reaction with Ferric Chloride: Ferric chloride (FeCl3) in water is acidic due to the hydrolysis of the Fe3+ ion. The increased pH caused by methylamine promotes the precipitation of hydrated ferric oxide (Fe(OH)3), which is often represented as Fe2O3.xH2O.

(iv) 

• Protonation of Aniline: During nitration, the strongly acidic conditions protonate the aniline, forming the anilinium ion (C6H5NH3+). The positively charged -NH3+ group is meta-directing.
• Competition: While some of the aniline remains as the free base and undergoes ortho- and para-nitration, the significant amount of anilinium ion formed leads to a substantial amount of meta-nitroaniline.

(v) 

• Salt Formation: The amino group (-NH2) of aniline reacts with the Lewis acid catalyst (e.g., AlCl3) used in the Friedel-Crafts reaction to form a salt. This salt formation deactivates the benzene ring, making it less susceptible to electrophilic substitution. The amino group becomes a strongly meta directing group.

(vi) 

• Resonance Stabilization: The diazonium ion of an aromatic amine is resonance-stabilized due to the delocalization of the positive charge into the aromatic ring. This delocalization makes the aromatic diazonium ion more stable.
• Less Resonance in Aliphatic: Aliphatic diazonium ions lack this resonance stabilization, making them considerably less stable and prone to rapid decomposition (loss of N2).

(vii) 

• Selective Formation of Primary Amines: The Gabriel phthalimide synthesis specifically produces primary amines without any secondary or tertiary amine byproducts. This is a significant advantage, as it avoids the difficult separation of amine mixtures.
• Mild Conditions: The reaction proceeds under relatively mild conditions.
• No Rearrangement: The reaction does not involve carbocation intermediates, so it avoids the possibility of rearrangements that can occur in some other amine synthesis methods.

4. Arrange the following:

(i) In decreasing order of pKb values:

C2H5NH2,C6H5NHCH3,(C2H5)2NH and C6H5NH2

(ii) In increasing order of basic strength:

C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2.

(iii) In increasing order of basic strength:

(а)Aniline,p-nitroaniline andp-toluidine

(b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2

(iv) In decreasing order of basic strength in gas phase:

C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3

(v) In increasing order of boiling point:

C2H5OH, (CH3)2NH, C2H5NH2

(vi) In increasing order of solubility in water:

C6H5NH2,(C2H5)2NH,C2H5NH2

Ans : 

(i) Decreasing order of pKb values:

A higher pKb value indicates a weaker base. Aliphatic amines are generally stronger bases than aromatic amines. Within aliphatic amines, secondary amines are usually more basic than primary or tertiary amines in aqueous solutions (due to a combination of factors including hydration and steric effects). Aryl groups attached to the nitrogen decrease basicity due to resonance.

1. C6H5NH2 (Aniline – aromatic amine, weakest base)
2. C6H5NHCH3 (N-Methylaniline – aromatic amine, but the methyl group increases electron density slightly compared to aniline)
3. C2H5NH2 (Ethylamine – primary aliphatic amine)
4. (C2H5)2NH (Diethylamine – secondary aliphatic amine, strongest base in this series)

(ii) Increasing order of basic strength:

1. C6H5NH2 (Aniline)
2. C6H5N(CH3)2 (N,N-Dimethylaniline – electron-donating methyl groups increase basicity compared to aniline)
3. CH3NH2 (Methylamine)
4. (C2H5)2NH (Diethylamine)

(iii) Increasing order of basic strength:

(a) Aniline, p-nitroaniline, and p-toluidine:

• Electron-donating groups (like -CH3 in p-toluidine) increase basicity.
• Electron-withdrawing groups (like -NO2 in p-nitroaniline) decrease basicity.

1. p-Nitroaniline
2. Aniline
3. p-Toluidine

(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2:

• Aliphatic amines are generally stronger bases than aromatic amines.

1. C6H5NH2 (Aniline)
2. C6H5NHCH3 (N-Methylaniline)
3. C6H5CH2NH2 (Benzylamine – the amino group is not directly attached to the benzene ring, so it behaves more like an aliphatic amine)

(iv) 

In the gas phase, steric hindrance is the dominant factor.

1. (C2H5)3N (Triethylamine – most sterically hindered)
2. (C2H5)2NH (Diethylamine)
3. C2H5NH2 (Ethylamine)
4. NH3 (Ammonia – least hindered)

(v) Increasing order of boiling point:

Boiling point depends on intermolecular forces. Alcohols can form strong hydrogen bonds, followed by primary amines (less strong H-bonds), and then secondary amines (even weaker H-bonds due to steric hindrance).

1. (CH3)2NH (Dimethylamine – weakest intermolecular forces)
2. C2H5NH2 (Ethylamine)
3. C2H5OH (Ethanol – strongest hydrogen bonding)

(vi) Increasing order of solubility in water:

Smaller amines are more soluble. Aryl groups decrease water solubility.

1. C6H5NH2 (Aniline – least soluble due to the large hydrophobic phenyl group)
2. (C2H5)2NH (Diethylamine – can form H-bonds, but less polar than primary amines)
3. C2H5NH2 (Ethylamine – most soluble due to smaller size and ability to form H-bonds)

5. How will you convert:

(i) Ethanoic acid into methanamine

(ii) Hexanenitrile into 1-aminopentane

(iii) Methanol to ethanoic acid.

(iv) Ethanamine into methanamine

(v) Ethanoic acid into propanoic acid

(vi) Methanamine into ethanamine

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid?

Ans : 

6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Ans : 

Let’s describe two common methods for distinguishing between primary, secondary, and tertiary amines, along with the relevant chemical equations.

1. Hinsberg’s Test:

This test relies on the reaction of amines with p-toluenesulfonyl chloride (TsCl, also known as Hinsberg’s reagent) to form sulfonamides. The differing solubilities of these sulfonamides in aqueous alkali allow us to classify the amines.  

• Primary Amines (RNH₂): React with TsCl to form N-alkyl-p-toluenesulfonamides. These sulfonamides contain an acidic hydrogen on the nitrogen and thus dissolve in aqueous KOH to form a soluble salt. Upon acidification, the insoluble sulfonamide precipitates.

RNH₂ + TsCl (in pyridine) → RNH-Ts + HCl  (Formation of sulfonamide)

RNH-Ts + KOH(aq) → RNH-Ts⁻ K⁺ (Soluble salt)

RNH-Ts⁻ K⁺ + H⁺ → RNH-Ts (Insoluble sulfonamide)

• Secondary Amines (R₂NH): React with TsCl to form N,N-dialkyl-p-toluenesulfonamides. These sulfonamides lack an acidic hydrogen and are insoluble in aqueous KOH.

R₂NH + TsCl (in pyridine) → R₂N-Ts + HCl (Formation of sulfonamide)

R₂N-Ts + KOH(aq) → No reaction (Insoluble)

• Tertiary Amines (R₃N): Generally do not react with TsCl under the conditions of the Hinsberg test.  

R₃N + TsCl → No reaction

2. Reaction with Nitrous Acid (HNO₂):

This method exploits the distinct reactions of the three classes of amines with nitrous acid (HNO₂), which is generated in situ by reacting sodium nitrite (NaNO₂) with dilute hydrochloric acid (HCl).

• Primary Amines (RNH₂): React with HNO₂ to form unstable diazonium salts. These diazonium salts decompose rapidly, even at low temperatures, evolving nitrogen gas (effervescence) and forming alcohols (or other products depending on the reaction conditions). Aromatic primary amines, however, form relatively more stable diazonium salts at low temperatures (0-5°C).  

RNH₂ + HNO₂ → RN₂⁺Cl⁻ + H₂O  (Diazotization)

RN₂⁺Cl⁻ + H₂O → ROH + N₂ + HCl (Decomposition of diazonium salt)

ArNH₂ + HNO₂ (0-5°C) → ArN₂⁺Cl⁻ (Aromatic diazonium salt)

• Secondary Amines (R₂NH): React with HNO₂ to form N-nitrosamines (also known as nitrosamides), which are often oily liquids and are potentially carcinogenic.

R₂NH + HNO₂ → R₂N-NO + H₂O (N-Nitrosamine formation)

• Tertiary Amines (R₃N): React with HNO₂ to form trialkylammonium nitrites. These salts are soluble in aqueous solution and usually produce no visible change in the reaction mixture.

R₃N + HNO₂ → R₃NH⁺NO₂⁻ (Trialkylammonium nitrite salt formation)

7. Write short notes on the following:

(i) Carbylamine reaction

(ii) Diazotisation

(iii) ‘Hofmann’s bromamide reaction

(iv) Coupling reaction

(v) Ammonolysis

(vi) Acetylation

(vii) Gabriel phthalimide synthesis

Ans : 

(i) Carbylamine Reaction (Isocyanide Test):

• Description: A chemical test for primary amines. A primary amine (aliphatic or aromatic) reacts with chloroform (CHCl₃) and potassium hydroxide (KOH) in alcoholic solution to form an isocyanide (carbylamine) which has a foul, unpleasant odor. Secondary and tertiary amines do not respond to this test.  

Reaction:

RNH₂ + CHCl₃ + 3KOH (alcoholic) → RNC + 3KCl + 3H₂O

• Significance: Used to detect the presence of a primary amine group.  

(ii) Diazotization:

• Description: The process of converting a primary aromatic amine (ArNH₂) into a diazonium salt (ArN₂⁺X⁻) by reaction with nitrous acid (HNO₂) at low temperatures (0-5°C). Nitrous acid is generated in situ from sodium nitrite (NaNO₂) and a mineral acid (e.g., HCl).  

Reaction:

ArNH₂ + NaNO₂ + 2HCl → ArN₂⁺Cl⁻ + NaCl + 2H₂O

• Significance: Diazonium salts are versatile intermediates in organic synthesis, used to introduce various substituents into aromatic rings.  

(iii) Hofmann Bromamide Reaction:

• Description: A method for preparing primary amines from amides. An amide is treated with bromine (Br₂) in an aqueous solution of KOH or NaOH. The reaction involves the loss of one carbon atom as CO₂ and yields a primary amine with one fewer carbon than the starting amide.  

Reaction:

RCONH₂ + Br₂ + 4KOH → RNH₂ + K₂CO₃ + 2KBr + H₂O

• Significance: Useful for synthesizing primary amines, particularly when other methods might be less effective.  

(iv) Coupling Reaction:

• Description: A reaction between a diazonium salt and a highly activated aromatic ring (such as in phenols or aromatic amines) to form an azo compound. The reaction occurs at the para position (or ortho if para is blocked) of the activated ring.

Reaction:

ArN₂⁺Cl⁻ + C₆H₅OH → Ar-N=N-C₆H₄-OH (p-Hydroxyphenylazo compound)

• Significance: Azo compounds are often brightly colored and are used as dyes.  

(v) Ammonolysis:

• Description: The reaction of alkyl halides with ammonia (NH₃) or an excess of an amine to form a mixture of primary, secondary, and tertiary amines, and quaternary ammonium salts.

Reaction:

RX + NH₃ → RNH₂ + HX

RNH₂ + RX → R₂NH + HX

R₂NH + RX → R₃N + HX

R₃N + RX → R₄N⁺X⁻

• Significance: A general method for preparing amines, although it often yields a mixture of products.
  

(vi) Acetylation:

• Description: The introduction of an acetyl group (CH₃CO-) into a molecule. This is commonly done by reacting alcohols, phenols, or amines with acetic anhydride (Ac₂O) or acetyl chloride (CH₃COCl), often in the presence of a base catalyst (like pyridine).

Reaction (with an amine):

RNH₂ + (CH₃CO)₂O → RNHCOCH₃ + CH₃COOH

• Significance: Used to protect amino groups during synthesis, or to modify the properties of organic molecules.  

(vii) Gabriel Phthalimide Synthesis:

• Description: A method for the synthesis of primary amines from alkyl halides. Phthalimide is converted to its potassium salt, which then reacts with the alkyl halide. Hydrolysis of the resulting N-alkylphthalimide with aqueous acid or base yields the primary amine.  

Reaction:

Phthalimide + KOH → Potassium phthalimide

Potassium phthalimide + RX → N-Alkylphthalimide

N-Alkylphthalimide + H₂O/H⁺ (or OH⁻) → RNH₂ + Phthalic acid

• Significance: A preferred method for preparing primary amines because it avoids the formation of secondary and tertiary amines (which are difficult to separate).

8. Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

(ii) Benzene to m-bromophenol

(iii) Benzoic acid to aniline

(iv) Aniline to 2,4,6-tribromofluorobenzene

(v) Benzyl chloride to 2-phenylethanamine

(vi) Chlorobenzene to p-Chloroaniline

(vii) Aniline to p-bromoaniIine

(viii)Benzamide to toluene

(ix) Aniline to benzyl alcohol.

Ans : 

9. Give the structures of A,B and C in the following reaction:

Ans : 

10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.

Ans:

Compound C: The molecular formula C₆H₇N strongly suggests an aromatic primary amine, likely aniline (C₆H₅NH₂). This is a key clue.

Compound B: Compound B is converted to aniline (C) by reaction with Br₂ and KOH. This is the Hofmann bromamide degradation reaction. This reaction converts an amide (RCONH₂) to a primary amine (RNH₂). Therefore, B must be benzamide (C₆H₅CONH₂).

Compound A: Compound A reacts with aqueous ammonia and heating to form benzamide (B). This is a reaction of a carboxylic acid derivative. Since the reaction involves aqueous ammonia, A is most likely benzoyl chloride (C₆H₅COCl).

11. Complete the following reactions:

Ans : 

12.  Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Ans : 

The Gabriel phthalimide synthesis relies on the phthalimide anion effectively attacking an organic halide through a nucleophilic substitution (SN2) mechanism. However, aryl halides are generally unreactive in such substitutions. Consequently, this crucial step cannot occur with aryl halides, and therefore, the Gabriel phthalimide synthesis is not a viable method for preparing aromatic primary amines (arylamines).

13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Ans :

14. Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans : 

(i) 

• Acidity and Proton Donation: Acidity refers to the ability of a compound to donate a proton (H⁺). Both alcohols (ROH) and amines (RNH₂) contain hydrogen atoms that could potentially be donated.
  
• Electronegativity: Oxygen is much more electronegative than nitrogen. This means that in an alcohol (ROH), the oxygen atom pulls electron density away from the hydrogen atom, making it easier for the hydrogen to be released as a proton (H⁺). In an amine (RNH₂), the nitrogen is less electronegative, so it holds onto the hydrogen more tightly.  
• Conjugate Base Stability: When an alcohol loses a proton, it forms an alkoxide ion (RO⁻). The negative charge on the oxygen is relatively stable due to oxygen’s high electronegativity. When an amine loses a proton, it forms an amide ion (RNH⁻). The negative charge on the nitrogen is less stable because nitrogen is less electronegative. A more stable conjugate base favors the release of the proton, hence greater acidity.  

Therefore, alcohols are more acidic than amines because the O-H bond is more polarized than the N-H bond, and the resulting conjugate base (alkoxide) is more stable than the amide ion.  

(ii) 

• Intermolecular Forces: Boiling point is related to the strength of intermolecular forces between molecules.  
• Hydrogen Bonding: Primary amines (RNH₂) have two hydrogen atoms directly bonded to the nitrogen. These hydrogen atoms can participate in strong hydrogen bonding with other primary amine molecules. Tertiary amines (R₃N) have no hydrogen atoms directly bonded to the nitrogen, so they cannot form hydrogen bonds with each other.  
• Other Factors: While dipole-dipole interactions and London dispersion forces are present in both primary and tertiary amines, the hydrogen bonding in primary amines is a much stronger intermolecular force.
  

Therefore, the presence of strong hydrogen bonding in primary amines leads to significantly higher boiling points compared to tertiary amines, where only weaker van der Waals forces (dipole-dipole and London dispersion forces) are present.  

(iii)

• Basicity and Proton Acceptance: Basicity refers to the ability of a compound to accept a proton (H⁺). Amines accept protons by using the lone pair of electrons on the nitrogen atom.  
• Electron Availability: In aliphatic amines, the nitrogen’s lone pair of electrons is readily available to accept a proton. Alkyl groups attached to the nitrogen are electron-donating (through the +I inductive effect), which increases the electron density on the nitrogen, making it even more ready to accept a proton.
• Delocalization in Aromatic Amines: In aromatic amines (like aniline), the nitrogen’s lone pair of electrons is delocalized into the aromatic ring through resonance.
  This delocalization decreases the electron density on the nitrogen atom, making it less available to accept a proton.